Probability and Statistical Analysis: Homeowner Meetings, Scam Outcomes, and Random Variables

QUESTION

2. A Nairobi couty political aspirant, is looking ahead to next year’s election campaign and the possibility that she might be forced to support a property tax increase to help combat projected shortfalls in the county budget. For the dual purpose of gaining voter support and finding out more about what her constituents think about taxes and other important issues, she is planning to hold a town meeting with a simple random sample of 40 homeowners from her county. For the more than 100,000 homes in Shirley’s county, the mean value is KES 19,000,000 and the standard deviation is KES 5,000,000. What is the probability that the mean value of the homes owned by those invited to her meeting will be;

1. greater than KES 20,000,000?
2. between KES 18,000,000 and KES 24,000,000?

1. Scam artists sometimes “stage” accidents by purposely walking in front of slow-moving luxury cars whose drivers are presumed to carry large amounts of liability insurance. The “victim” then offers to accept a small monetary cash settlement so the driver can avoid an insurance claim, a blot on his driving record, and increased insurance premiums.

Marlin has just tried to stage such a scheme by walking into a slow-moving Mercedes and faking a minor injury. Assume that the possible outcomes for Marlin are KES 50000 (i.e., he is arrested and fined for his scam), KES 0 (the driver recognizes the scam and refuses to pay), and KES 10000 (the driver simply pays up to get rid of Marlin). If the probabilities for these outcomes are 0.1, 0.3, and 0.6, respectively, what is Marlin’s expected monetary outcome for trying to scam the Mercedes driver?

Indicate whether each of the following random variables is discrete or continuous.

1. The diameter of aluminum rods coming off a production line
2. The number of years of schooling employees have completed
3. The Dow Jones Industrial Average
4. The volume of milk purchased during a supermarket visit

The average Kenyan family of four spends KES 500000 per year on food prepared at home. Assuming a normal distribution with a standard deviation of KES 100000 and a randomly selected Kenyan family of four, what is the probability that the family’s annual spending for food prepared at home will be:

1. more than KES 800000?
2. between KES 500000 and KES 700000?
3. less than KES 600000?

ANSWER

 Probability and Statistical Analysis: Homeowner Meetings, Scam Outcomes, and Random Variables

Introduction

In the context of a Nairobi county political aspirant preparing for an election campaign, various statistical and probability concepts are being explored. This essay will delve into the probabilities associated with homeowner meetings, the expected outcome of a scam attempt, and the nature of certain random variables.

Probability in Homeowner Meetings

The political aspirant is organizing a town meeting with a simple random sample of 40 homeowners from the county. The mean value of homes in the county is KES 19,000,000, with a standard deviation of KES 5,000,000. Two specific probabilities are of interest:

Probability of the mean home value being greater than KES 20,000,000:
To calculate this probability, we need to find the z-score for the value of KES 20,000,000, given the mean and standard deviation. The formula for the z-score is (X – μ) / σ, where X is the value, μ is the mean, and σ is the standard deviation. Substituting the values, we find the z-score and use a standard normal distribution table or calculator to find the probability.

Probability of the mean home value being between KES 18,000,000 and KES 24,000,000:
Similar to the previous calculation, we find the z-scores for both values and then find the corresponding probabilities. The probability between these two values can be found by subtracting the probability of the lower value from the probability of the higher value.

Expected Outcome of the Scam Attempt

In the scenario involving Marlin’s attempt to stage a scam with potential outcomes of KES 50,000, KES 0, and KES 10,000, with probabilities 0.1, 0.3, and 0.6 respectively, we can calculate Marlin’s expected monetary outcome. The expected value is calculated by multiplying each outcome by its probability and summing up the results.

Nature of Random Variables

Certain variables are characterized as either discrete or continuous. The examples provided are:

The diameter of aluminum rods coming off a production line: Continuous variable, as it can take on any value within a certain range.

The number of years of schooling employees have completed: Discrete variable, as it takes on distinct and separate values (e.g., 10 years, 12 years).

The Dow Jones Industrial Average: Continuous variable, as it can take on any value within a certain range, including fractions.

The volume of milk purchased during a supermarket visit: Continuous variable, as it can take on any value within a certain range.

Probability of Family’s Annual Spending on Food Prepared at Home

Assuming a normal distribution of annual spending on food prepared at home, with a mean of KES 500,000 and a standard deviation of KES 100,000, the probabilities of various spending ranges can be calculated using z-scores and the standard normal distribution table or calculator. The same approach as used earlier can be applied to find the probabilities of spending more than KES 800,000, between KES 500,000 and KES 700,000, and less than KES 600,000.

Conclusion

In the world of statistical analysis and probability, these concepts play a pivotal role in decision-making, understanding distributions, and making informed predictions. By applying these concepts to scenarios like homeowner meetings, scam outcomes, and the nature of random variables, individuals can better grasp the underlying trends and uncertainties within their environments, aiding in making well-informed choices.